For every integer n, which of the following expressions is always even?

n^2 + n

Correct answerYour answer

n^2 + 1

Correct answerYour answer

3n + 1

Correct answerYour answer

n^3 + 2

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$n^2 + n = n(n + 1)$ , a product of two consecutive integers.
One of any two consecutive integers is even, so the product is always even.
The others fail for some n: $n^2 + 1 = 5$ at $n = 2$ , $3n + 1 = 7$ at $n = 2$ , $n^3 + 2 = 3$ at $n = 1$ .

Worth noting

Testing only one value is the trap: $n^2 + 1$ is even when n is odd, but not when n is even.

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Explanation